Show that A is nonsingular if and only if 0 is not an eigenv

Show that A is nonsingular if and only if 0 is not an eigenvalue of A. (Thus we could add another line to the list in Theorem 1.2.3.)

Solution

Let, A be nonsingular

Assume, 0 is an eigenvalue

Hence, det(A-0*I)=det(A)=0

Hence, A is singular which is a contracition

Hence 0 is not an eigenvalue

Let, 0 not be an eigenvalue

Assuming, A is singular

det(A)=0=det(A-0*I)

Hence, 0 is an eigenvalue

which is a contradictino

Hence, A is non singular